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.99999999... converges to 1, and for all real-world applications is equal to 1.
However, there is the question of whether 1/infinity (the number that .9999... would have to be added to to equal 1) is equal to zero or not is really the point we are debating.
I say that since the numerator is a non-zero term, that it is not actually equal to zero, no matter how large the denominator is.
Of course, since 1/infinity isn't a real number (since infinity is also not a real number), it just gets messy.
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I think your a little confused.
".99999999... converges to 1, and for all real-world applications is equal to 1. "
No , this is wrong. You state converges to, but how can you talk about converges in respect of a single number. A finite summation may converge as the number of terms increases to infinity. But 0.9r is not a finite summation it is a symbolic represntation which exactly equals one. We dont say 0.3r converges to 1/3 , convergance doesnt come into it, it is exactly equal to 1/3. To talk about convergance you must provide an ordered measure.
In summary you can say f(n)=9*sum(0.1^i,i=1,n) converges to 0.9r as n(in N) tends to infinity. 0.9r exacty is 1, in real world or the mathematical world.
Quote:
However, there is the question of whether 1/infinity (the number that .9999... would have to be added to to equal 1) is equal to zero or not is really the point we are debating.
I say that since the numerator is a non-zero term, that it is not actually equal to zero, no matter how large the denominator is.
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Here you make the mistake of accepting the notation of 1/infinity as valid. It is not valid or correct notation, it makes no sense.
In such cases it does pay to consider convergance. Consider
y=sin(x)/x
at x = 0, we have y=0/0 which is undefined. But ofcourse by the limit as x tends to zero is 1 by l'hopitals rule .
See if you can work out what the limit is as x tends to infinity
